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Introductory Chemistry, 3 rd Edition Nivaldo Tro Chapter 6 Chemical Composition Roy Kennedy Massachusetts Introductory Chemistry, 3 rd Edition Nivaldo Tro Chapter 6 Chemical Composition Roy Kennedy Massachusetts Bay Community College Wellesley Hills, MA 2009, Prentice Hall

Why Is Knowledge of Composition Important? • All matter is either chemically or physically Why Is Knowledge of Composition Important? • All matter is either chemically or physically combined into substances. • Knowing the fraction of material you have can tell you: ü the amount of sodium in sodium chloride for diet. ü the amount of iron in iron ore for steel production. ü the amount of hydrogen in water for hydrogen fuel. ü the amount of chlorine in freon to estimate ozone depletion. Tro's "Introductory Chemistry", Chapter 6 2

How much seed do you plant? • In a garden you count the seeds How much seed do you plant? • In a garden you count the seeds by hand. How many seeds would you know to plant in a field? 3

Counting by Weighing • Building a house requires a lot of nails. • If Counting by Weighing • Building a house requires a lot of nails. • If you know that a single nail weighs. 0122 g, than 100 nails weigh 1. 22 g, a 1000 nails weigh 12. 2 g and so on. • Analogy: üYou want to make 100 lbs of Al 2 O 3, how much aluminum do you use Tro's "Introductory Chemistry", Chapter 6 4

Counting Nails by the Pound, Continued A hardware store customer buys 2. 60 pounds Counting Nails by the Pound, Continued A hardware store customer buys 2. 60 pounds of nails. A dozen nails has a mass of 0. 150 pounds. How many nails did the customer buy? 1 dozen nails = 0. 150 lbs. 12 nails = 1 dozen nails Solution map: Tro's "Introductory Chemistry", Chapter 6 5

Counting Nails by the Pound, Continued • The customer bought 2. 60 lbs of Counting Nails by the Pound, Continued • The customer bought 2. 60 lbs of nails and received 208 nails. He counted the nails by weighing them! Tro's "Introductory Chemistry", Chapter 6 6

Counting Nails by the Pound, Continued • What if he bought a different size Counting Nails by the Pound, Continued • What if he bought a different size nail? üWould the mass of a dozen be 0. 150 lbs? üWould there still be 12 nails in a dozen? üWould there be 208 nails in 2. 60 lbs? üHow would this effect the conversion factors? Tro's "Introductory Chemistry", Chapter 6 7

Counting Atoms by Moles • If we can find the mass of a particular Counting Atoms by Moles • If we can find the mass of a particular number of atoms, we can use this information to convert the mass of an element sample to the number of atoms in the sample. • The number of atoms we will use is 6. 022 x 1023 and we call this a mole. ü 1 mole = 6. 022 x 1023 things. ØLike 1 dozen = 12 things. üAvogadro’s number. üLike a kilo = 1000 or a Google = 1× 10100 Tro's "Introductory Chemistry", Chapter 6 8

Chemical Packages—Moles • Mole = Number of carbon atoms “in” 12 g of C-12. Chemical Packages—Moles • Mole = Number of carbon atoms “in” 12 g of C-12. ü 1 mole protons or 1 mole of neutrons = 1 amu üC-12 exactly 6 protons and 6 neutrons üsince 1 mole × 1 amu = 1 g. ü 1 mole of C-12 (which is 12 amu) weighs exactly 12 g. • In 12 g of C-12 there are 6. 022 x 1023 C-12 atoms. Tro's "Introductory Chemistry", Chapter 6 9

Example 6. 1: • A silver ring contains 1. 1 x 1022 silver atoms. Example 6. 1: • A silver ring contains 1. 1 x 1022 silver atoms. How many moles of silver are in the ring? Tro's "Introductory Chemistry", Chapter 6 10

Example: A silver ring contains 1. 1 x 1022 silver atoms. How many moles Example: A silver ring contains 1. 1 x 1022 silver atoms. How many moles of silver are in the ring? • Write down the given quantity and its units. Given: 1. 1 x 1022 Ag atoms Tro's "Introductory Chemistry", Chapter 6 11

Example: A silver ring contains 1. 1 x 1022 silver atoms. How many moles Example: A silver ring contains 1. 1 x 1022 silver atoms. How many moles of silver are in the ring? Information: Given: 1. 1 x 1022 Ag atoms • Write down the quantity to find and/or its units. Find: ? moles Tro's "Introductory Chemistry", Chapter 6 12

Example: A silver ring contains 1. 1 x 1022 silver atoms. How many moles Example: A silver ring contains 1. 1 x 1022 silver atoms. How many moles of silver are in the ring? Information: Given: 1. 1 x 1022 Ag atoms Find: ? moles • Collect needed conversion factors: 1 mole Ag atoms = 6. 022 x 1023 Ag atoms. Tro's "Introductory Chemistry", Chapter 6 13

Example: A silver ring contains 1. 1 x 1022 silver atoms. How many moles Example: A silver ring contains 1. 1 x 1022 silver atoms. How many moles of silver are in the ring? Information: Given: 1. 1 x 1022 Ag atoms Find: ? moles Conversion Factor: 1 mole = 6. 022 x 1023 • Write a solution map for converting the units: atoms Ag Tro's "Introductory Chemistry", Chapter 6 moles Ag 14

Example: A silver ring contains 1. 1 x 1022 silver atoms. How many moles Example: A silver ring contains 1. 1 x 1022 silver atoms. How many moles of silver are in the ring? Information: Given: 1. 1 x 1022 Ag atoms Find: ? moles Conversion Factor: 1 mole = 6. 022 x 1023 Solution Map: atoms mole • Apply the solution map: = 1. 8266 x 10 -2 moles Ag • Significant figures and round: = 1. 8 x 10 -2 moles Ag Tro's "Introductory Chemistry", Chapter 6 15

Example: A silver ring contains 1. 1 x 1022 silver atoms. How many moles Example: A silver ring contains 1. 1 x 1022 silver atoms. How many moles of silver are in the ring? Information: Given: 1. 1 x 1022 Ag atoms Find: ? moles Conversion Factor: 1 mole = 6. 022 x 1023 Solution Map: atoms mole • Check the solution: 1. 1 x 1022 Ag atoms = 1. 8 x 10 -2 moles Ag The units of the answer, moles, are correct. The magnitude of the answer makes sense since 1. 1 x 1022 is less than 1 mole. Tro's "Introductory Chemistry", Chapter 6 16

Practice—Calculate the Number of Atoms in 2. 45 Mol of Copper. Tro's Practice—Calculate the Number of Atoms in 2. 45 Mol of Copper. Tro's "Introductory Chemistry", Chapter 6 17

Practice—Calculate the Number of Atoms in 2. 45 Mol of Copper, Continued. Given: Find: Practice—Calculate the Number of Atoms in 2. 45 Mol of Copper, Continued. Given: Find: 2. 45 mol Cu atoms Cu Solution Map: mol Cu Relationships: 1 mol = 6. 022 x 1023 atoms Cu Solution: Check: Since atoms are small, the large number of atoms makes sense. Tro's "Introductory Chemistry", Chapter 6 18

Relationship Between Moles and Mass • The mass of one mole of atoms is Relationship Between Moles and Mass • The mass of one mole of atoms is called the molar mass. • The molar mass of an element, in grams, is numerically equal to the element’s atomic mass, in amu. • The lighter the atom, the less a mole weighs. • The lighter the atom, the more atoms there are in 1 g. Tro's "Introductory Chemistry", Chapter 6 19

Mole and Mass Relationships 1 mole sulfur 32. 06 g 1 mole carbon 12. Mole and Mass Relationships 1 mole sulfur 32. 06 g 1 mole carbon 12. 01 g Tro's "Introductory Chemistry", Chapter 6 20

Example 6. 2—Calculate the Moles of Sulfur in 57. 8 g of Sulfur. Given: Example 6. 2—Calculate the Moles of Sulfur in 57. 8 g of Sulfur. Given: Find: Solution Map: 57. 8 g S mol S g. S mol S Relationships: 1 mol S = 32. 07 g Solution: Check: Since the given amount is much less than 1 mol S, the number makes sense. Tro's "Introductory Chemistry", Chapter 6 21

Practice—Calculate the Moles of Carbon in 0. 0265 g of Pencil Lead. Tro's Practice—Calculate the Moles of Carbon in 0. 0265 g of Pencil Lead. Tro's "Introductory Chemistry", Chapter 6 22

Practice—Calculate the Moles of Carbon in 0. 0265 g of Pencil Lead, Continued. Given: Practice—Calculate the Moles of Carbon in 0. 0265 g of Pencil Lead, Continued. Given: Find: Solution Map: 0. 0265 g C mol C g. C mol C Relationships: 1 mol C = 12. 01 g Solution: Check: Since the given amount is much less than 1 mol C, the number makes sense. Tro's "Introductory Chemistry", Chapter 6 23

Example 6. 3: • How many aluminum atoms are in an aluminum can with Example 6. 3: • How many aluminum atoms are in an aluminum can with a mass of 16. 2 g? Tro's "Introductory Chemistry", Chapter 6 24

Example 6. 3—How Many Aluminum Atoms Are in a Can Weighing 16. 2 g? Example 6. 3—How Many Aluminum Atoms Are in a Can Weighing 16. 2 g? Given: Find: Solution Map: Relationships: 16. 2 g Al atoms Al g Al mol Al atoms Al 1 mol Al = 26. 98 g, 1 mol = 6. 022 x 1023 Solution: Check: Since the given amount is much less than 1 mol Cu, the number makes sense. Tro's "Introductory Chemistry", Chapter 6 25

Molar Mass of Compounds • The relative weights of molecules can be calculated from Molar Mass of Compounds • The relative weights of molecules can be calculated from atomic weights. Formula mass = 1 molecule of H 2 O = 2(1. 01 amu H) + 16. 00 amu O = 18. 02 amu. • Since 1 mole of H 2 O contains 2 moles of H and 1 mole of O. Molar mass = 1 mole H 2 O = 2(1. 01 g H) + 16. 00 g O = 18. 02 g. Tro's "Introductory Chemistry", Chapter 6 26

Example 6. 4—Calculate the Mass of 1. 75 Mol of H 2 O. Given: Example 6. 4—Calculate the Mass of 1. 75 Mol of H 2 O. Given: Find: Solution Map: 1. 75 mol H 2 O g H 2 O Relationships: 1 mol H 2 O = 18. 02 g Solution: Check: Since the given amount is more than 1 mol, the mass being > 18 g makes sense. Tro's "Introductory Chemistry", Chapter 6 27

Practice—How Many Moles Are in 50. 0 g of Pb. O 2? (Pb = Practice—How Many Moles Are in 50. 0 g of Pb. O 2? (Pb = 207. 2, O = 16. 00) Tro's "Introductory Chemistry", Chapter 6 28

Practice—How Many Moles Are in 50. 0 g of Pb. O 2? (Pb = Practice—How Many Moles Are in 50. 0 g of Pb. O 2? (Pb = 207. 2, O = 16. 00), Continued Given: Find: Solution Map: 50. 0 g mol Pb. O 2 moles Pb. O 2 g Pb. O 2 mol Pb. O 2 Relationships: 1 mol Pb. O 2 = 239. 2 g Solution: Check: Since the given amount is less than 239. 2 g, the moles being < 1 makes sense. Tro's "Introductory Chemistry", Chapter 6 29

Example 6. 5—What Is the Mass of 4. 78 x 1024 NO 2 Molecules? Example 6. 5—What Is the Mass of 4. 78 x 1024 NO 2 Molecules? Given: 4. 78 x 1024 NO 2 molecules Find: g NO 2 Solution Map: molecules Relationships: mol NO 2 g NO 2 1 mol NO 2 = 46. 01 g, 1 mol = 6. 022 x 1023 Solution: Check: Since the given amount is more than Avogadro’s number, the mass > 46 g makes sense. Tro's "Introductory Chemistry", Chapter 6 30

Counting and ratio’s It takes me. 2 gal of gas to get to IVC. Counting and ratio’s It takes me. 2 gal of gas to get to IVC. It is a very simple ratio: = What if I only had. 1 gal 200 2 X 4’s, 3 sinks, 2 showers, you can make a house with 3 bathrooms and 3 bedrooms. What if you had 12 sinks…how many houses could you make. 200 3 2 = 1 3 3

Chemical Formulas as Conversion Factors • 1 spider 8 legs. • 1 chair 4 Chemical Formulas as Conversion Factors • 1 spider 8 legs. • 1 chair 4 legs. • 1 H 2 O molecule 2 H atoms 1 O atom. Tro's "Introductory Chemistry", Chapter 6 32

Mole Relationships in Chemical Formulas • Since we count atoms and molecules in mole Mole Relationships in Chemical Formulas • Since we count atoms and molecules in mole units, we can find the number of moles of a constituent element if we know the number of moles of the compound. Moles of compound 1 mol Na. Cl 1 mol H 2 O 1 mol Ca. CO 3 Moles of constituents 1 mol Na, 1 mol Cl 2 mol H, 1 mol O 1 mol Ca, 1 mol C, 3 mol O 1 mol C 6 H 12 O 6 6 mol C, 12 mol H, 6 mol O Tro's "Introductory Chemistry", Chapter 6 33

Example 6. 6—Calculate the Moles of Oxygen in 1. 7 Moles of Ca. CO Example 6. 6—Calculate the Moles of Oxygen in 1. 7 Moles of Ca. CO 3. Given: Find: Solution Map: 1. 7 mol Ca. CO 3 mol O Relationships: 1 mol Ca. CO 3 = 3 mol O Solution: Check: Since the given amount is much less than 1 mol S, the number makes sense. Tro's "Introductory Chemistry", Chapter 6 34

Example 6. 7: • Carvone (C 10 H 14 O) is the main component Example 6. 7: • Carvone (C 10 H 14 O) is the main component in spearmint oil. It has a pleasant odor and mint flavor. It is often added to chewing gum, liqueurs, soaps, and perfumes. Find the mass of carbon in 55. 4 g of carvone. Tro's "Introductory Chemistry", Chapter 6 35

Example: Find the mass of carbon in 55. 4 g of carvone, (C 10 Example: Find the mass of carbon in 55. 4 g of carvone, (C 10 H 14 O). • Write down the given quantity and its units. Given: 55. 4 g C 10 H 14 O Tro's "Introductory Chemistry", Chapter 6 36

Example: Find the mass of carbon in 55. 4 g of carvone, (C 10 Example: Find the mass of carbon in 55. 4 g of carvone, (C 10 H 14 O). Information: Given: 55. 4 g C 10 H 14 O • Write down the quantity to find and/or its units. Find: ? g C Tro's "Introductory Chemistry", Chapter 6 37

Example: Find the mass of carbon in 55. 4 g of carvone, (C 10 Example: Find the mass of carbon in 55. 4 g of carvone, (C 10 H 14 O). Information: Given: 55. 4 g C 10 H 14 O Find: g C • Collect needed conversion factors: Molar mass C 10 H 14 O = 10(atomic mass C) + 14(atomic mass H) + 1(atomic mass O) = 10(12. 01) + 14(1. 01) + (16. 00) = 150. 2 g/mol 1 mole C 10 H 14 O = 150. 2 g C 10 H 14 O 1 mole C 10 H 14 O 10 mol C 1 mole C = 12. 01 g C Tro's "Introductory Chemistry", Chapter 6 38

Example: Find the mass of carbon in 55. 4 g of carvone, (C 10 Example: Find the mass of carbon in 55. 4 g of carvone, (C 10 H 14 O). Information: Given: 55. 4 g C 10 H 14 O Find: g C Conversion Factors: 1 mol C 10 H 14 O = 150. 2 g 1 mol C 10 H 14 O 10 mol C 1 mol C = 12. 01 g • Write a solution map for converting the units: g C 10 H 14 O mol C 10 H 14 O Tro's "Introductory Chemistry", Chapter 6 mol C g C 39

Example: Find the mass of carbon in 55. 4 g of carvone, (C 10 Example: Find the mass of carbon in 55. 4 g of carvone, (C 10 H 14 O). Information: Given: 55. 4 g C 10 H 14 O Find: g C Conversion Factors: 1 mol C 10 H 14 O = 150. 2 g 1 mol C 10 H 14 O 10 mol C 1 mol C = 12. 01 g Solution Map: g C 10 H 14 O mol C g C • Apply the solution map: = 44. 2979 g C • Significant figures and round: = 44. 3 g C Tro's "Introductory Chemistry", Chapter 6 40

Percent Composition • Percentage of each element in a compound. ü By mass. • Percent Composition • Percentage of each element in a compound. ü By mass. • Can be determined from: ü The formula of the compound. ü The experimental mass analysis of the compound. • The percentages may not always total to 100% due to rounding. Tro's "Introductory Chemistry", Chapter 6 42

Example 6. 9—Find the Mass Percent of Cl in C 2 Cl 4 F Example 6. 9—Find the Mass Percent of Cl in C 2 Cl 4 F 2. Given: Find: C 2 Cl 4 F 2 % Cl by mass Solution Map: Relationships: Solution: Check: Since the percentage is less than 100 and Cl is much heavier than the other atoms, the number makes sense.

Practice—Determine the Mass Percent Composition of the Following: Ca. Cl 2 (Ca = 40. Practice—Determine the Mass Percent Composition of the Following: Ca. Cl 2 (Ca = 40. 08, Cl = 35. 45) Tro's "Introductory Chemistry", Chapter 6 44

Practice—Determine the Percent Composition of the Following, Continued: Ca. Cl 2 Tro's Practice—Determine the Percent Composition of the Following, Continued: Ca. Cl 2 Tro's "Introductory Chemistry", Chapter 6 45

Mass Percent as a Conversion Factor • The mass percent tells you the mass Mass Percent as a Conversion Factor • The mass percent tells you the mass of a constituent element in 100 g of the compound. üThe fact that Na. Cl is 39% Na by mass means that 100 g of Na. Cl contains 39 g Na. • This can be used as a conversion factor. ü 100 g Na. Cl 39 g Na Tro's "Introductory Chemistry", Chapter 6 46

Empirical Formulas • The simplest, whole-number ratio of atoms in a molecule is called Empirical Formulas • The simplest, whole-number ratio of atoms in a molecule is called the empirical formula. üCan be determined from percent composition or combining masses. • The molecular formula is a multiple of the empirical formula. Tro's "Introductory Chemistry", Chapter 6 47

Empirical (CH 2 O) -vs- Molecular Formula molecular formula Formaldehyde CH 2 O Acetic Empirical (CH 2 O) -vs- Molecular Formula molecular formula Formaldehyde CH 2 O Acetic Acid C 2 H 4 O 2 Lactic Acid C 3 H 6 O 3 Erythrose C 4 H 8 O 4 Ribose C 5 H 10 O 5 Glucose C 6 H 12 O 6 48 1 2 3 4 5 6 30. 03 g/mol 60. 06 g/mol 90. 09 g/mol 120. 12 g/mol 150. 15 g/mol 180. 18 g/mol

Empirical Formulas, Continued Hydrogen Peroxide Molecular formula = H 2 O 2 Empirical formula Empirical Formulas, Continued Hydrogen Peroxide Molecular formula = H 2 O 2 Empirical formula = HO Benzene Molecular formula = C 6 H 6 Empirical formula = CH Glucose Molecular formula = C 6 H 12 O 6 Empirical formula = CH 2 O 49

Example 6. 11—Finding an Empirical Formula from Experimental Data Tro's Example 6. 11—Finding an Empirical Formula from Experimental Data Tro's "Introductory Chemistry", Chapter 6 50

Example: • A laboratory analysis of aspirin determined the following mass percent composition. Find Example: • A laboratory analysis of aspirin determined the following mass percent composition. Find the empirical formula. C = 60. 00% H = 4. 48% O = 35. 53% Tro's "Introductory Chemistry", Chapter 6 51

Example: Find the empirical formula of aspirin with the given mass percent composition. • Example: Find the empirical formula of aspirin with the given mass percent composition. • Write down the given quantity and its units. Given: C = 60. 00% H = 4. 48% O = 35. 53% Therefore, in 100 g of aspirin there are 60. 00 g C, 4. 48 g H, and 35. 53 g O. Tro's "Introductory Chemistry", Chapter 6 52

Example: Find the empirical formula of aspirin with the given mass percent composition. Information: Example: Find the empirical formula of aspirin with the given mass percent composition. Information: Given: 60. 00 g C, 4. 48 g H, 35. 53 g O • Write down the quantity to find and/or its units. Find: empirical formula, Cx. Hy. Oz Tro's "Introductory Chemistry", Chapter 6 53

Example: Find the empirical formula of aspirin with the given mass percent composition. Information: Example: Find the empirical formula of aspirin with the given mass percent composition. Information: Given: 60. 00 g C, 4. 48 g H, 35. 53 g O Find: empirical formula, Cx. Hy. Oz • Collect needed conversion factors: 1 mole C = 12. 01 g C 1 mole H = 1. 01 g H 1 mole O = 16. 00 g O Tro's "Introductory Chemistry", Chapter 6 54

Example: Find the empirical formula of aspirin with the given mass percent composition. Information: Example: Find the empirical formula of aspirin with the given mass percent composition. Information: Given: 60. 00 g C, 4. 48 g H, 35. 53 g O Find: empirical formula, Cx. Hy. Oz Conversion Factors: 1 mol C = 12. 01 g; 1 mol H = 1. 01 g; 1 mol O = 16. 00 g • Write a solution map: g. C mol C g. H mol H g. O pseudoformula mole ratio whole number ratio empirical formula mol O Tro's "Introductory Chemistry", Chapter 6 55

Example: Find the empirical formula of aspirin with the given mass percent composition. Information: Example: Find the empirical formula of aspirin with the given mass percent composition. Information: Given: 60. 00 g C, 4. 48 g H, 35. 53 g O Find: empirical formula, Cx. Hy. Oz Conversion Factors: 1 mol C = 12. 01 g; 1 mol H = 1. 01 g; 1 mol O = 16. 00 g Solution Map: g C, H, O mol ratio empirical formula • Apply the solution map: ü Calculate the moles of each element. Tro's "Introductory Chemistry", Chapter 6 56

Example: Find the empirical formula of aspirin with the given mass percent composition. Information: Example: Find the empirical formula of aspirin with the given mass percent composition. Information: Given: 4. 996 mol C, 4. 44 mol H, 2. 221 mol O Find: empirical formula, Cx. Hy. Oz Conversion Factors: 1 mol C = 12. 01 g; 1 mol H = 1. 01 g; 1 mol O = 16. 00 g Solution Map: g C, H, O mol ratio empirical formula • Apply the solution map: ü Write a pseudoformula. C 4. 996 H 4. 44 O 2. 221 Tro's "Introductory Chemistry", Chapter 6 57

Example: Find the empirical formula of aspirin with the given mass percent composition. Information: Example: Find the empirical formula of aspirin with the given mass percent composition. Information: Given: C 4. 996 H 4. 44 O 2. 221 Find: empirical formula, Cx. Hy. Oz Conversion Factors: 1 mol C = 12. 01 g; 1 mol H = 1. 01 g; 1 mol O = 16. 00 g Solution Map: g C, H, O mol ratio empirical formula • Apply the solution map: ü Find the mole ratio by dividing by the smallest number of moles. Tro's "Introductory Chemistry", Chapter 6 58

Information: Given: C 2. 25 H 2 O 1 Find: empirical formula, Cx. Hy. Information: Given: C 2. 25 H 2 O 1 Find: empirical formula, Cx. Hy. Oz Conversion Factors: 1 mol C = 12. 01 g; 1 mol H = 1. 01 g; 1 mol O = 16. 00 g Solution Map: g C, H, O mol ratio empirical formula Example: Find the empirical formula of aspirin with the given mass percent composition. • Apply the solution map: ü Multiply subscripts by factor to give whole number. { } x 4 C 9 H 8 O 4 Tro's "Introductory Chemistry", Chapter 6 59

Example 6. 12—Finding an Empirical Formula from Experimental Data Tro's Example 6. 12—Finding an Empirical Formula from Experimental Data Tro's "Introductory Chemistry", Chapter 6 60

Example: • A 3. 24 -g sample of titanium reacts with oxygen to form Example: • A 3. 24 -g sample of titanium reacts with oxygen to form 5. 40 g of the metal oxide. What is the formula of the oxide? Tro's "Introductory Chemistry", Chapter 6 61

Example: Find the empirical formula of oxide of titanium with the given elemental analysis. Example: Find the empirical formula of oxide of titanium with the given elemental analysis. • Write down the given quantity and its units. Given: Ti = 3. 24 g compound = 5. 40 g Tro's "Introductory Chemistry", Chapter 6 62

Example: Find the empirical formula of oxide of titanium with the given elemental analysis. Example: Find the empirical formula of oxide of titanium with the given elemental analysis. Information: Given: 3. 24 g Ti, 5. 40 g compound • Write down the quantity to find and/or its units. Find: empirical formula, Tix. Oy Tro's "Introductory Chemistry", Chapter 6 63

Example: Find the empirical formula of oxide of titanium with the given elemental analysis. Example: Find the empirical formula of oxide of titanium with the given elemental analysis. Information: Given: 3. 24 g Ti, 5. 40 g compound Find: empirical formula, Tix. Oy • Collect needed conversion factors: 1 mole Ti = 47. 88 g Ti 1 mole O = 16. 00 g O Tro's "Introductory Chemistry", Chapter 6 64

Example: Find the empirical formula of oxide of titanium with the given elemental analysis. Example: Find the empirical formula of oxide of titanium with the given elemental analysis. Information: Given: 3. 24 g Ti, 5. 40 g compound Find: empirical formula, Tix. Oy Conversion Factors: 1 mol Ti = 47. 88 g; 1 mol O = 16. 00 g • Write a solution map: g Ti mol Ti pseudoformula g. O mole ratio whole number ratio empirical formula mol O Tro's "Introductory Chemistry", Chapter 6 65

Information: Given: 3. 24 g Ti, 5. 40 g compound Find: empirical formula, Tix. Information: Given: 3. 24 g Ti, 5. 40 g compound Find: empirical formula, Tix. Oy Conversion Factors: 1 mol Ti= 47. 88 g; 1 mol O= 16. 00 g Solution Map: g Ti, O mol ratio empirical formula Example: Find the empirical formula of oxide of titanium with the given elemental analysis. • Apply the solution map: ü Calculate the mass of each element. 5. 40 g compound − 3. 24 g Ti = 2. 16 g O Tro's "Introductory Chemistry", Chapter 6 66

Information: Given: 3. 24 g Ti, 2. 16 g O Find: empirical formula, Tix. Information: Given: 3. 24 g Ti, 2. 16 g O Find: empirical formula, Tix. Oy Conversion Factors: 1 mol Ti= 47. 88 g; 1 mol O= 16. 00 g Solution Map: g Ti, O mol ratio empirical formula Example: Find the empirical formula of oxide of titanium with the given elemental analysis. • Apply the solution map: ü Calculate the moles of each element. Tro's "Introductory Chemistry", Chapter 6 67

Example: Find the empirical formula of oxide of titanium with the given elemental analysis. Example: Find the empirical formula of oxide of titanium with the given elemental analysis. Information: Given: 0. 0677 mol Ti, 0. 135 mol O Find: empirical formula, Tix. Oy Conversion Factors: 1 mol Ti= 47. 88 g; 1 mol O= 16. 00 g Solution Map: g Ti, O mol ratio empirical formula • Apply the solution map: ü Write a pseudoformula. Ti 0. 0677 O 0. 135 Tro's "Introductory Chemistry", Chapter 6 68

Example: Find the empirical formula of oxide of titanium with the given elemental analysis. Example: Find the empirical formula of oxide of titanium with the given elemental analysis. Information: Given: 0. 0677 mol Ti, 0. 135 mol O Find: empirical formula, Tix. Oy Conversion Factors: 1 mol Ti= 47. 88 g; 1 mol O= 16. 00 g Solution Map: g Ti, O mol ratio empirical formula • Apply the solution map: ü Find the mole ratio by dividing by the smallest number of moles. Tro's "Introductory Chemistry", Chapter 6 69

Practice—Determine the Empirical Formula of Tin (II) Fluoride, which Contains 75. 7% Sn (118. Practice—Determine the Empirical Formula of Tin (II) Fluoride, which Contains 75. 7% Sn (118. 70) and the Rest Fluorine (19. 00). Tro's "Introductory Chemistry", Chapter 6 70

Practice—Determine the Empirical Formula of Tin (II) Fluoride, which Contains 75. 7% Sn (118. Practice—Determine the Empirical Formula of Tin (II) Fluoride, which Contains 75. 7% Sn (118. 70) and the Rest Fluorine (19. 00), Continued. Given: 75. 7% Sn, (100 – 75. 3) = 24. 3% F in 100 g Tin (II) fluoride there are 75. 7 g Sn and 24. 3 g F. Find: Snx. Fy Conversion Factors: 1 mol Sn = 118. 70 g; 1 mol F = 19. 00 g Solution Map: g Sn mol Sn g. F mol F pseudoformula Tro's "Introductory Chemistry", Chapter 6 whole mole number ratio empirical formula 71

Practice—Determine the Empirical Formula of Tin (II) Fluoride, which Contains 75. 7% Sn (118. Practice—Determine the Empirical Formula of Tin (II) Fluoride, which Contains 75. 7% Sn (118. 70) and the Rest Fluorine (19. 00), Continued. Apply solution map: Sn 0. 638 F 1. 28 Sn. F 2 Tro's "Introductory Chemistry", Chapter 6 72

Practice—Determine the Empirical Formula of Hematite, which Contains 72. 4% Fe (55. 85) and Practice—Determine the Empirical Formula of Hematite, which Contains 72. 4% Fe (55. 85) and the Rest Oxygen (16. 00). Tro's "Introductory Chemistry", Chapter 6 73

Practice—Determine the Empirical Formula of Hematite, which Contains 72. 4% Fe (55. 85) and Practice—Determine the Empirical Formula of Hematite, which Contains 72. 4% Fe (55. 85) and the Rest Oxygen (16. 00), Continued. Given: 72. 4% Fe, (100 – 72. 4) = 27. 6% O in 100 g hematite there are 72. 4 g Fe and 27. 6 g O. Find: Fex. Oy Conversion Factors: 1 mol Fe = 55. 85 g; 1 mol O = 16. 00 g Solution Map: whole g Fe mol Fe g. O mol O pseudoformula Tro's "Introductory Chemistry", Chapter 6 mole number ratio empirical formula 74

Practice—Determine the Empirical Formula of Hematite, which Contains 72. 4% Fe (55. 85) and Practice—Determine the Empirical Formula of Hematite, which Contains 72. 4% Fe (55. 85) and the Rest Oxygen (16. 00), Continued. Apply solution map: Fe 1. 30 O 1. 73 Tro's "Introductory Chemistry", Chapter 6 75

All These Molecules Have the Same Empirical Formula. How Are the Molecules Different? Name All These Molecules Have the Same Empirical Formula. How Are the Molecules Different? Name Glyceraldehyde Molecular Formula C 3 H 6 O 3 Empirical Formula CH 2 O Erythrose C 4 H 8 O 4 CH 2 O Arabinose C 5 H 10 O 5 CH 2 O Glucose C 6 H 12 O 6 CH 2 O Tro's "Introductory Chemistry", Chapter 6 76

All These Molecules Have the Same Empirical Formula. How Are the Molecules Different? , All These Molecules Have the Same Empirical Formula. How Are the Molecules Different? , Continued Name Glyceraldehyde Molecular Formula C 3 H 6 O 3 Empirical Formula CH 2 O (30) Molar Mass, g 90 Erythrose C 4 H 8 O 4 CH 2 O 120 Arabinose C 5 H 10 O 5 CH 2 O 150 Glucose C 6 H 12 O 6 CH 2 O 180 Tro's "Introductory Chemistry", Chapter 6 77

Molecular Formulas • The molecular formula is a multiple of the empirical formula. • Molecular Formulas • The molecular formula is a multiple of the empirical formula. • To determine the molecular formula, you need to know the empirical formula and the molar mass of the compound. Molar massreal formula = Factor used to multiply subscripts Molar massempirical formula Tro's "Introductory Chemistry", Chapter 6 78

Example—Determine the Molecular Formula of Cadinene if it has a Molar Mass of 204 Example—Determine the Molecular Formula of Cadinene if it has a Molar Mass of 204 g and an Empirical Formula of C 5 H 8. 1. Determine the empirical formula. ü May need to calculate it as previous. C 5 H 8 2. Determine the molar mass of the empirical formula. 5 C = 60. 05, 8 H = 8. 064 C 5 H 8 = 68. 11 g/mol Tro's "Introductory Chemistry", Chapter 6 79

Example—Determine the Molecular Formula of Cadinene if it has a Molar Mass of 204 Example—Determine the Molecular Formula of Cadinene if it has a Molar Mass of 204 g and an Empirical Formula of C 5 H 8, Continued. 3. Divide the given molar mass of the compound by the molar mass of the empirical formula. ü Round to the nearest whole number. Tro's "Introductory Chemistry", Chapter 6 80

Example—Determine the Molecular Formula of Cadinene if it has a Molar Mass of 204 Example—Determine the Molecular Formula of Cadinene if it has a Molar Mass of 204 g and an Empirical Formula of C 5 H 8, Continued. 4. Multiply the empirical formula by the factor above to give the molecular formula. (C 5 H 8)3 = C 15 H 24 Tro's "Introductory Chemistry", Chapter 6 81

Practice—Benzopyrene has a Molar Mass of 252 g and an Empirical Formula of C Practice—Benzopyrene has a Molar Mass of 252 g and an Empirical Formula of C 5 H 3. What is its Molecular Formula? (C = 12. 01, H=1. 01) Tro's "Introductory Chemistry", Chapter 6 82

Practice—Benzopyrene has a Molar Mass of 252 g and an Empirical Formula of C Practice—Benzopyrene has a Molar Mass of 252 g and an Empirical Formula of C 5 H 3. What is its Molecular Formula? (C = 12. 01, H=1. 01), Continued C 5 = 5(12. 01 g) = 60. 05 g H 3 = 3(1. 01 g) = 3. 03 g C 5 H 3 = 63. 08 g Molecular formula = {C 5 H 3} x 4 = C 20 H 12 Tro's "Introductory Chemistry", Chapter 6 83

Practice—Determine the Molecular Formula of Nicotine, which has a Molar Mass of 162 g Practice—Determine the Molecular Formula of Nicotine, which has a Molar Mass of 162 g and is 74. 0% C, 8. 7% H, and the Rest N. (C=12. 01, H=1. 01, N=14. 01) Tro's "Introductory Chemistry", Chapter 6 84

Practice—Determine the Molecular Formula of Nicotine, which has a Molar Mass of 162 g Practice—Determine the Molecular Formula of Nicotine, which has a Molar Mass of 162 g and is 74. 0% C, 8. 7% H, and the Rest N, Continued Given: 74. 0% C, 8. 7% H, {100 – (74. 0+8. 7)} = 17. 3% N in 100 g nicotine there are 74. 0 g C, 8. 7 g H, and 17. 3 g N. Find: Cx. Hy. Nz Conversion Factors: 1 mol C = 12. 01 g; 1 mol H = 1. 01 g; 1 mol N = 14. 01 g Solution Map: g. C mol C g. H mol H g. N mol N pseudoformula Tro's "Introductory Chemistry", Chapter 6 whole mole number ratio empirical formula 85

Practice—Determine the Molecular Formula of Nicotine, which has a Molar Mass of 162 g Practice—Determine the Molecular Formula of Nicotine, which has a Molar Mass of 162 g and is 74. 0% C, 8. 7% H, and the Rest N, Continued. Apply solution map: C 5 = 5(12. 01 g) = N 1 = 1(14. 01 g) = H 7 = 7(1. 01 g) = C 5 H 7 N = 60. 05 g 14. 01 g 7. 07 g 81. 13 g C 6. 16 H 8. 6 N 1. 23 {C 5 H 7 N} x 2 = C 10 H 14 N 2 Tro's "Introductory Chemistry", Chapter 6 86